P(Not sibling) = 1 - P(sibling)

Let the siblings be (AB) (CD) (EFG)

Picking up AB - 1 possibility
Picking up CD - 1 possibility
Picking up 2 from EFG - 3C2 - 3 possibilities

Total = 5.

Total Number of ways of picking up from 7 - 7C2 = 21.

1 - 5/21 = 16/21. E IMO

The wording can throw you off, but basically I picture this group of 7 as follows: There are 7 people (A, B, C, D, E, F, and G) split into 2 groups. 1 group of 4 people has exactly 1 sibling in the room. This means that there are 2 pairs of siblings. We will assume A & B are siblings and C & D are siblings. The secong group of 3 people has exactly 2 siblings in the group - each other. So E, F, and G are siblings with each other. We want to find the probability that two people selected are NOT siblings.

First we can find the total number of ways to pick two people out of a group of 7. Since the order does not matter, it is 7!/(5!*2!) or 21 different ways. There is 1 way that each of the 2 pairs of siblings can be selected and 3 ways that the trio of siblings can be selected. This is a total of 5 ways siblings can be selected.

Since this is a very small group, we can list the ways that siblings can be selected:

- A&B
- C&D
- E&F
- F&G
- E&G

There are 5 ways siblings can be selected so there must be 21-5 = 16 ways where siblings are not selected. So the answer is 16/21 or E.

Its all about rephrasing the question!! Thanks a lot CappyAA